Find the inverse of the matrix, $\text B = \left[\begin{array}{rr}3 & 6 \\ -10 & -10\end{array}\right]$. Non-integers should be given either as decimals or as simplified fractions. $ B^{-1}=$
Explanation: The Strategy To find the inverse of an invertible matrix, we can use Gaussian Elimination. To do this, we do the following. First, we append the matrix $\text B$ with the identity matrix $\text I$ to get [ B | I ] \left[\begin{array} ~\text B ~ |~\text I\end{array}\right]. Next, we use Gaussian Elimination to reduce $\text B$ to the identity matrix, $\text I$. Performing the same operations on $\text I$ will convert it to $\text B^{-1}$, so that our new matrix becomes [ I | B − 1 ] \left[\begin{array} ~\text I ~ |~\text B^{-1}\end{array}\right]. Appending $\text B$ with $\text I$ [ B | I ] = [ 3 − 10 6 − 10 1 0 0 1 ] \left[\begin{array} ~\text B ~ |~\text I\end{array}\right]=\left[\begin{array}{rr}3 & 6 & 1 & 0 \\ -10 & -10 & 0 & 1 \end{array}\right] Eliminating the leading term in the second row We want the first term of $R_2$ to equal $0$, so we add $\dfrac{10}{3}R_1$ to $R_2$. $\left[\begin{array}{rr}3 & 6 & 1 & 0 \\ {-10} & {-10} & {0} & {1} \end{array}\right]\xrightarrow{R_2+\dfrac{10}{3}R_1\rightarrow R_2}\left[\begin{array}{rr}3 & 6 & 1 & 0 \\ {0} & {10} & {\dfrac{10}{3}} & {1} \end{array}\right]$ Reducing the leading terms and back-solving Now, let's reduce the leading term of $R_2$ to equal $1$. $\left[\begin{array}{rr}3 & 6 & 1 & 0 \\ {0} & {10} & {\dfrac{10}{3}} & {1} \end{array}\right]\xrightarrow{\dfrac{1}{10}R_2\rightarrow R_2}\left[\begin{array}{rr}3 & 6 & 1 & 0 \\ {0} & {1} & {\dfrac{1}{3}} & {\dfrac{1}{10}} \end{array}\right]$ We are ready to back-solve to get [ I | B − 1 ] \left[\begin{array} ~\text I ~ |~\text B^{-1}\end{array}\right]. $\begin{aligned}\!\!\left[\begin{array}{rr}\!{3}\! & {6}\! & {1}\! & {0} \\ \!0\! & 1\! & \dfrac{1}{3}\! & \dfrac{1}{10} \!\end{array}\right]\!\xrightarrow{\!\!R_1-6R_2\rightarrow R_1\!\!} \!\!&\left[\begin{array}{rr}\!{3}\!\! & {0}\!\! & {-1} \!\!& {-\dfrac{3}{5}} \!\\ \!0\! & 1\! & \dfrac{1}{3}\! & \dfrac{1}{10} \end{array}\right] \!\!\xrightarrow{\!\!\dfrac{1}{3}R_1\rightarrow R_1\!\!}\!\!\left[\begin{array}{rr}{1}\!\! & {0}\!\! & {-\dfrac{1}{3}}\!\! & {-\dfrac{1}{5}} \\\!0\! & 1\! & \dfrac{1}{3}\! & \dfrac{1}{10} \end{array}\right]\end{aligned}$ Therefore $\text B^{-1}=\left[\begin{array}{rr} -\dfrac{1}{3}\!\! & -\dfrac{1}{5} \\ \dfrac{1}{3}\! & \dfrac{1}{10} \end{array}\right]$. Summary $\text B^{-1}=\left[\begin{array}{rr} -\dfrac{1}{3}\!\! & -\dfrac{1}{5} \\ \dfrac{1}{3}\! & \dfrac{1}{10} \end{array}\right]$